Grade 11 : Physics - Motion (Straight/Plane) (Set 3)

Explanation: Velocity $v(t) = 3t(t-2)$ changes sign at $t = 2\text{ s}$.
Displacement from $t=0$ to 2: $s_1 = \int_0^2 (3t^2-6t)dt = [t^3-3t^2]_0^2 = -4m$ (distance = 4 m).
Displacement from $t=2$ to 3: $s_2 = [t^3-3t^2]_2^3 = 0 - (-4) = +4m$ (distance = 4 m).
Total distance $= 4 + 4 = 8m$.

Explanation: Maximum height $H = \frac{u^2 \sin^2\theta}{2g}$ and range $R = \frac{2u^2 \sin\theta \cos\theta}{g}$.
Setting $H = R$ yields $\frac{u^2 \sin^2\theta}{2g} = \frac{2u^2 \sin\theta \cos\theta}{g}$.
Simplifying gives $\tan\theta = 4 \implies \theta = \tan^{-1}(4)$.

Explanation: Relative speed of approach $v_{rel} = 50 + 70 = 120km/h$.
Time taken $t = \frac{\text{Distance}}{v_{rel}} = \frac{240km}{120km/h} = 2.0hours$.

Explanation: Speed at $t = 10\text{ s}$ is $v = a_t t = 3 \times 10 = 30m/s$.
Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{900}{100} = 9m/s^2$.
Net acceleration $a = \sqrt{a_t^2 + a_c^2} = \sqrt{3^2 + 9^2} = \sqrt{90} = 3\sqrt{10}m/s^2$.

Explanation: For shortest crossing time, the swimmer points perpendicular to the bank.
Time taken $t = \frac{d}{v_s} = \frac{120}{5} = 24\text{ s}$.
Downstream drift $x = v_r \times t = 3 \times 24 = 72m$.

Explanation: Velocity crosses zero at $t = 2\text{ s}$.
Area under $v-t$ curve from 0 to $2\text{ s} = \frac{1}{2} \times 2 \times 10 = 10m$.
Area from 2 to $4\text{ s} = \frac{1}{2} \times 2 \times |-10| = 10m$.
Total distance is the sum of magnitudes $= 10 + 10 = 20m$.

Explanation: Displacement of A: $y_A = \frac{1}{2}gt^2$ downwards.
Displacement of B: $y_B = ut - \frac{1}{2}gt^2$ upwards.
Sum of distances: $y_A + y_B = H \implies \frac{1}{2}gt^2 + ut - \frac{1}{2}gt^2 = H \implies ut = H \implies t = \frac{H}{u}$.

Explanation: Velocity vector $\vec{v}(t) = \frac{d\vec{r}}{dt} = 8t \hat{i} + 3 \hat{j}$.
At $t = 2\text{ s}$, $\vec{v}(2) = 16 \hat{i} + 3 \hat{j}$.
Speed $|\vec{v}| = \sqrt{16^2 + 3^2} = \sqrt{256 + 9} = \sqrt{265}m/s$.

Explanation: At maximum height, vertical component of velocity $v_y = 0$.
Horizontal velocity remains constant: $v_x = u \cos 60^\circ = 20 \times 0.5 = 10m/s$.
Hence velocity vector is $10\hat{i}m/s$.

Explanation: $v(t) = \int (6t-2)dt = 3t^2 - 2t + C_1$. Since $v(0)=4$, $v(t) = 3t^2 - 2t + 4$.
$x(t) = \int (3t^2-2t+4)dt = t^3 - t^2 + 4t + C_2$. Since $x(0)=0$, $x(t) = t^3 - t^2 + 4t$.
At $t = 2\text{ s}$, $x(2) = 2^3 - 2^2 + 4(2) = 8 - 4 + 8 = 12m$.

Explanation: Displacement of Car A: $x_A = 20t$.
Displacement of Car B: $x_B = \frac{1}{2}(4)t^2 = 2t^2$.
For Car B to overtake Car A: $2t^2 = 20t \implies 2t(t - 10) = 0 \implies t = 10\text{ s}$.

Explanation: Compare with standard trajectory formula $y = x \tan\theta - \frac{g x^2}{2u^2 \cos^2\theta}$.
Here $\tan\theta = \sqrt{3} \implies \theta = 60^\circ$, so $\cos\theta = 0.5$.
Equating coefficient of $x^2$: $\frac{g}{2u^2 (0.25)} = \frac{g}{20} \implies 0.5 u^2 = 20 \implies u^2 = 40 \implies u = 2\sqrt{10}m/s$.

Explanation: For equal distances, average speed is the harmonic mean of speeds:
$v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{2 \times 30 \times 60}{30 + 60} = \frac{3600}{90} = 40km/h$.

Explanation: Relative velocity of rain w.r.t man $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m = -4\hat{j} - 3\hat{i}$.
Angle $\theta$ with vertical satisfies $\tan\theta = \frac{v_m}{v_r} = \frac{3}{4}$.
Therefore $\theta = \tan^{-1}(3/4)$.

Explanation: Distance covered in $n$-th second is $D_n = u + \frac{g}{2}(2n - 1) = \frac{g}{2}(2n - 1)$ for $u=0$.
For $n = 1, 2, 3$, ratio of $(2n-1)$ is $1 : 3 : 5$. (Galileo's law of odd numbers).

Explanation: Magnitude of change in velocity for speed $v$ turning through angle $\theta$ is given by $|\Delta \vec{v}| = 2v \sin(\theta/2)$.
For $\theta = 60^\circ$, $|\Delta \vec{v}| = 2v \sin(30^\circ) = 2v \left(\frac{1}{2}\right) = v$.

Explanation: Velocity $v(t) = \frac{dx}{dt} = 5 - 4t$.
Instantaneous rest occurs when $v(t) = 0 \implies 5 - 4t = 0 \implies t = \frac{5}{4} = 1.25\text{ s}$.

Explanation: Range $R_1 = \frac{u^2 \sin(2(45^\circ - \theta))}{g} = \frac{u^2 \sin(90^\circ - 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g}$.
Range $R_2 = \frac{u^2 \sin(2(45^\circ + \theta))}{g} = \frac{u^2 \sin(90^\circ + 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g}$.
Since $R_1 = R_2$, the ratio is $1 : 1$.

Explanation: Velocity vector $\vec{v}_A = 40\hat{i}$, $\vec{v}_B = 30\hat{j}$.
Relative velocity $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B = 40\hat{i} - 30\hat{j}$.
Speed $= \sqrt{40^2 + (-30)^2} = 50km/h$. Direction has positive East (+i) and negative North (-j, i.e., South), so South-East.

Explanation: From $v^2 = u^2 + 2as \implies 0 = u^2 - 2ad \implies d = \frac{u^2}{2a}$.
Stopping distance is directly proportional to the square of initial speed ($d \propto u^2$).
If speed is tripled ($3v$), new stopping distance $= (3)^2 d = 9d$.